Theorem: Every non-constant polynomial with complex coefficients has a zero in $\mathbb{C}$.

Proof: Let $P(z)$ be any non-constant polynomial. Assume by way of contradiction that $P(z)\neq 0$ for all $z\in\mathbb{C}$. Then the function $f(z):=1/P(z)$ is an entire function. Since $P$ is non-constant, $P\rightarrow \infty$ as $z\rightarrow\infty$, so that $f$ is bounded. By Liouville's Theorem $f$ is constant. Then $P$ is constant, a contradiction. Hence $P$ has a zero in $\mathbb{C}$.

I am trying to understand this proof, but I am stuck on "$P\rightarrow\infty$ as $z\rightarrow\infty$." Could somebody provide further details, please?

  • 2
    $\begingroup$ $P(z)=z^n+\ldots $, with leading term $z^n$. Show that this means $P\to \infty$ for $z\to \infty$. $\endgroup$ – Dietrich Burde Feb 23 '15 at 20:53

Assume that $P(z) = a_nz^n + a_{n-1}z^{n-1} + \dots + a_1 z + a_0$. Then $P(z) = z^n \left( a_n + \frac{a_{n-1}}{z} + \dots + \frac{a_1}{z^{n-1}} + \frac{a_0}{z^n}\right)$, and so if $z\to \infty$, you might notice something happens with the other terms.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.